Questions on conservation of energy

Question 1

State the principle of conservation of energy.

Accepted Answer

Energy cannot be created or destroyed, only transferred from one form to another (1); the total energy of a closed system stays constant (1).

Question 2

Describe the main energy transfer that takes place as a ball falls freely towards the ground.

Accepted Answer

Gravitational potential energy is transferred to kinetic energy (1) as the ball speeds up on the way down (1).

Question 3

Calculate the kinetic energy of a 4.0 kg object moving at 5.0 m s⁻¹.

Accepted Answer

KE = ½mv² = ½ × 4.0 × 5.0² (1) KE = 50 J (1)

Question 4

A pendulum bob swings from its highest point on one side, through the lowest point, to its highest point on the other side. Air resistance is negligible. Describe the energy transfers during this swing.

Accepted Answer

Gravitational potential energy is transferred to kinetic energy as the bob falls to the lowest point, where the kinetic energy is greatest (1); the kinetic energy is then transferred back to gravitational potential energy as the bob rises to the other side. (1)

Question 5

The speed of a car doubles. State the factor by which its kinetic energy changes, and justify your answer.

Accepted Answer

The kinetic energy increases by a factor of 4 (1), because KE = ½mv², so it is proportional to the square of the speed and 2² = 4. (1)

Question 6

State the condition under which the sum of an object's kinetic and gravitational potential energy stays constant as it falls.

Accepted Answer

Resistive forces (air resistance) must be negligible, so no work is done against resistance. (1)

Question 7

An object is dropped from rest and falls 1.8 m. Using energy conservation and taking g = 9.81 m s⁻², calculate its speed just before it lands (ignore air resistance).

Accepted Answer

mgh = ½mv² (1) v = √(2gh) = √(2 × 9.81 × 1.8) (1) v = 5.94 m s⁻¹ (1)

Question 8

A 0.20 kg ball is dropped from a height of 2.5 m. Taking g = 9.81 m s⁻², calculate (a) its gravitational potential energy at the start and (b) its speed just before impact.

Accepted Answer

(a) GPE = mgh = 0.20 × 9.81 × 2.5 (1) GPE = 4.9 J (1) (b) All of this becomes kinetic energy (1) v = √(2gh) = 7.00 m s⁻¹ (1)

Question 9

A ball leaves the ground moving vertically upwards at 6.0 m s⁻¹. Using energy conservation with g = 9.81 m s⁻², calculate the maximum height it reaches.

Accepted Answer

½mv² = mgh (1) h = v²/2g = 6.0²/(2 × 9.81) (1) h = 1.83 m (1)

Question 10

A 5.0 kg block slides 4.0 m down a slope inclined at 30°, against a constant frictional force of 10 N. Taking g = 9.81 m s⁻², use energy conservation to determine its kinetic energy and speed at the bottom.

Accepted Answer

Height drop = 4.0 sin 30° = 2.0 m (1)
GPE released = mgh = 5.0 × 9.81 × 2.0 = 98.1 J (1)
Work against friction = 10 × 4.0 = 40 J (1)
KE at bottom = 98.1 − 40 = 58.1 J, so v = √(2 × KE/m) = 4.82 m s⁻¹ (1)

Question 11

A light child and a heavy adult start from rest at the top of identical smooth slides. Explain, using energy conservation, why they reach the bottom with the same speed.

Accepted Answer

With no friction, mgΔh = ½mv² (1). The mass cancels, giving v = √(2gΔh) (1), which depends only on the height drop, and that is the same for both. (1)

Question 12

A cyclist and bicycle, of total mass 85 kg, travel at 12 m s⁻¹ along a level road. The brakes provide a constant total resistive force of 340 N. Using energy conservation, determine the distance travelled while coming to rest.

Accepted Answer

KE = ½ × 85 × 12² = 6120 J (1). All of it is transferred by the work done against the braking force: Fd = KE (1), so d = 6120/340 = 18 m. (1)

Question 13

A stone is thrown at 8.0 m s⁻¹ from a bridge 12 m above a river. Air resistance is negligible; g = 9.81 m s⁻². (a) Calculate the speed of the stone as it enters the water. (b) Explain why the answer does not depend on the direction in which the stone is thrown.

Accepted Answer

(a) ½mv² = ½mu² + mgΔh, so v² = u² + 2gΔh (1) = 8.0² + 2 × 9.81 × 12 = 299 m² s⁻² (1); v = 17.3 m s⁻¹ (1). (b) Energy is a scalar: the kinetic energy gained equals mgΔh whatever the direction of the throw, so the final speed is fixed by the height drop alone. (1)

Question 14

A roller-coaster car starts from rest at the top of a 25 m drop and passes a point 10 m above the ground. Ignoring friction and taking g = 9.81 m s⁻², calculate its speed at that point.

Accepted Answer

Height fallen = 25 − 10 = 15 m (1) ½mv² = mgΔh (1) v = √(2g × 15) (1) v = 17.2 m s⁻¹ (1)

Question 15

A ball is dropped from a height of 2.0 m and bounces back to a height of 1.5 m. Taking g = 9.81 m s⁻², calculate (a) the fraction of its energy retained in the bounce and (b) its speed just after leaving the ground.

Accepted Answer

(a) Energy ∝ height (1) Fraction retained = 1.5/2.0 = 0.75, that is 75% (1) (b) v = √(2g × 1.5) (1) v = 5.42 m s⁻¹ (1)

Question 16

Energy is always conserved, yet we still talk about 'wasting' energy. Explain this apparent contradiction.

Accepted Answer

The total energy is conserved (1), but useful energy is progressively transferred to dissipated forms such as heat and sound that are spread out and hard to recover (1). 'Wasted' energy means energy transferred to a less useful form, not energy lost (1).

Question 17

A skateboarder of mass 68 kg claims that her ramp is almost frictionless. She starts from rest at the top of the ramp, 3.2 m above the bottom, and a speed gate at the bottom records 6.0 m s⁻¹. Taking g = 9.81 m s⁻², deduce whether her claim is justified.

Accepted Answer

GPE given up = mgΔh = 68 × 9.81 × 3.2 = 2135 J (1). KE at the bottom = ½ × 68 × 6.0² = 1224 J (1). Energy transferred to the surroundings by resistive forces = 2135 − 1224 = 911 J (1), which is 911/2135 = 43% of the input (1). Nearly half the energy is dissipated, so the ramp is far from frictionless and her claim is wrong. (1)

Question 18

A pendulum bob of mass 0.15 kg is released from a point 80 mm above its lowest position. After 20 complete swings the bob rises only 30 mm above the lowest position. Taking g = 9.81 m s⁻², calculate the average energy dissipated per swing.

Accepted Answer

Initial GPE = mgΔh = 0.15 × 9.81 × 0.080 = 0.118 J (1). After 20 swings GPE = 0.15 × 9.81 × 0.030 = 0.044 J (1). Energy dissipated = 0.118 − 0.044 = 0.074 J (1). Average per swing = 0.074/20 = 3.7 × 10⁻³ J. (1)

Question 19

The car of a roller coaster is hauled by a motor to the top of the first hill, then released and runs unpowered over several lower hills before brakes bring it to rest at the end of the ride. Describe fully the energy transfers throughout the ride, and explain why each successive hill must be lower than the one before.

Accepted Answer

The motor does work on the car, transferring energy to its gravitational potential store (1). On each descent gravitational potential energy is transferred to kinetic energy, and on each climb kinetic energy is transferred back to gravitational potential energy (1). Throughout the ride, friction and air resistance transfer some energy to internal energy of the track, wheels and air (1). The total of the car's kinetic and potential energy therefore falls steadily, so the car cannot regain its previous height and each hill must be lower than the one before (1). The brakes transfer the remaining kinetic energy to internal energy (heating) (1). At every stage the total energy, including the dissipated energy, is conserved. (1)

Question 20

Both energy equations of this topic follow from the work done by a force, W = Fs. Derive ΔE_p = mgΔh, and then derive E_k = ½mv² using the equations of motion as well. State the condition each derivation needs.

Accepted Answer

To raise a mass m through a height Δh without giving it any speed, the applied force must equal the weight, mg, and it acts along the displacement, so W = Fs = mgΔh (1). That work is stored in the gravitational field and returned in full if the mass is released, so ΔE_p = mgΔh (1). Treating g as a single number requires the field to be uniform over the height climbed (1). For the kinetic energy, let a constant resultant force F accelerate a mass m from rest through a distance s: W = Fs = mas, using F = ma (1). With u = 0, v² = u² + 2as leaves as = ½v², so W = m × ½v² = ½mv², and nothing was lifted or rubbed, so every joule went into the motion (1). That step uses v² = u² + 2as, so the acceleration must be constant and the mass must start from rest (1).